How do you convert #2=(3x+7y)^2-x# into polar form?

1 Answer
Cem Sentin
Dec 17, 2017

#29r^2-20r^2cos(2theta)+21r^2sin(2theta)-rcos(theta)-2=0#

Explanation:

#2=(3x+7y)^2-x#

#9x^2+42xy+49y^2-x=2#

After using #x=rcos(theta)# and #y=rsin(theta)# transormation,

#9r^2(cos(theta))^2+42r^2*cos(theta)*sin(theta)+49r^2(sin(theta))^2-rcos(theta)=2#

#9r^2*(1+cos(2theta))/2+21r^2*sin(2theta)+49r^2(1-cos(2theta))/2-rcos(theta)=2#

#29r^2-20r^2cos(2theta)+21r^2sin(2theta)-rcos(theta)-2=0#

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